标签：wordList word cog Ladder transformation beginWord endWord Word leetcode
Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:
- Only one letter can be changed at a time.
- Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
Note:
- Return 0 if there is no such transformation sequence.
- All words have the same length.
- All words contain only lowercase alphabetic characters.
- You may assume no duplicates in the word list.
- You may assume beginWord and endWord are non-empty and are not the same.
Example 1:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: 5 Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", return its length 5.
Example 2:
Input: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log"] Output: 0 Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
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我发现我现在实力很弱，得更加努力了。
这个题是说的就是把一个字符串改变其中一个字符，并且保证改变后的字符串再给定的一个字典上有，然后，如果最后改得到的字符串时符合要求的字符串，统计改变的次数，否则返回0. 和图的遍历类似，只是每次要遍历26个字符，用BFS最合适。
参考链接：http://www.cnblogs.com/grandyang/p/4539768.html（强烈推荐这个大佬）
C++代码：
class Solution {
public:
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> wordset(wordList.begin(),wordList.end()); //建立一个字典。
if(!wordset.count(endWord)) return 0; //如果endWord不在字典上，那无论改多少次，都不能得到这个字符串。
queue<string> q;
q.push(beginWord);
int res = 0;
while(!q.empty()){
for(int k = q.size(); k > 0; k--){
string word = q.front(); q.pop();
if(word == endWord) return res + 1;
for(int i = 0; i < word.size(); i++){
string newword = word;
for(char ch = 'a'; ch <= 'z';ch++){
newword[i] = ch;
if(wordset.count(newword) && newword!=word){
q.push(newword);
wordset.erase(newword); //由于转变后得到的字符串不能与前面的相同，所以得在字典上删除相应的字符串。
}
}
}
}
++res;
}
return 0;
}
};
标签：wordList,word,cog,Ladder,transformation,beginWord,endWord,Word,leetcode 来源： https://www.cnblogs.com/Weixu-Liu/p/10812885.html
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